Question

In Problems 7–10 the independent variable x is missing in the given differential equation. Proceed as in Example 2 and solve the equation by using the substitution u = y'.

7.

$\mathit{y}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\left(y^{\text{'}}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$

Short answer

The solution is

$(x+c{)}^2+y^2=k$

Step-by-step solution

given information

$y{y}^{\text{'}\text{'}}+{\left(y^{\text{'}}\right)}^2+1=0$

solve the equation

We have the differential equation

$y{y}^{\text{'}\text{'}}+{\left(y^{\text{'}}\right)}^2+1=0....\left(1\right)$

and we have to solve it as the following technique

First, we have the substitution

y' = u .......(2)

After that, since we have y' = u , then by differentiation with respect to x, we can have

$$\begin{gathered} y^{\text{'}\text{'}}=\frac{du}{dx} \\ =\frac{du}{dy}\times \frac{dy}{dx} \\ =y^{\text{'}}\frac{du}{dy} \\ =u\frac{du}{dy}...\left(3\right) \end{gathered}$$

Second, substitute from equations (2) and (3) y' and y'' into the given equation (1), then we obtain

$$\begin{gathered} y\left(u\frac{du}{dy}\right)+(u{)}^2+1=0 \\ y\left(u\frac{du}{dy}\right)=-\left(1+u^2\right) \\ \frac{-u}{1+u^2}du=\frac{dy}{y} \end{gathered}$$

After that, we have to do integration as

$-\frac{1}{2}\int \frac{2udu}{1+u^2}=\int \frac{dy}{y}$

substitution

This integration $\int \frac{2udu}{1+u^2}$ can be made by substitution with $v=1+u^2,$then we have $dv=2udu$, then we obtain

$$\begin{gathered} -\frac{1}{2}\int \frac{dv}{v}=\int \frac{dy}{y} \\ -\frac{1}{2}lnv=lny+lnc \\ ln\sqrt{v}=-lny+ln{c}_1 \\ ln\sqrt{1+u^2}=ln\left(\frac{c_1}{y}\right) \\ {e}^{ln\sqrt{1+u^2}}=e^{ln\left(\frac{c_1}{y}\right)} \\ \sqrt{1+u^2}=\frac{c_1}{y} \\ 1+u^2=\frac{{\left(c_1\right)}^2}{y^2} \\ {u}^2=\frac{k-y^2}{y^2} \\ u=\sqrt{\frac{k-y^2}{y^2}}....\left(4\right) \end{gathered}$$

Third, substitute with equation (2) into equation (4), then we have

$$\begin{gathered} y^{\text{'}}=\sqrt{\frac{k-y^2}{y^2}} \\ \frac{dy}{dx}=\sqrt{\frac{k-y^2}{y^2}} \\ \frac{dy}{\sqrt{\frac{k-y^2}{y^2}}}=dx \\ \frac{\sqrt{y^2}dy}{\sqrt{k-y^2}}=dx \\ \frac{ydy}{\sqrt{k-y^2}}=dx \end{gathered}$$

substitute constants

After that, we have to do integration as

$$\begin{gathered} \int \frac{ydy}{\sqrt{k-y^2}}=\int dx \\ -\frac{1}{2}\int \frac{-2ydy}{\sqrt{k-y^2}}=\int dx \end{gathered}$$

This integration $\int \frac{-2ydy}{\sqrt{k-y^2}}$can be made by substitution with $s=k-y^2$ then we have $ds=-2ydy$, then we obtain

$$\begin{gathered} -\frac{1}{2}\int \frac{-2ydy}{\sqrt{k-y^2}}=\int dx \\ -\frac{1}{2}\int \frac{ds}{\sqrt{s}}=\int dx \\ -\frac{1}{2}\times 2\sqrt{s}=x+c \\ x+c=-\sqrt{s} \end{gathered}$$

Then we obtain

role="math" $\begin{array}{rcl}x+c& =& -\sqrt{k-y^2}\\ (x+c{)}^2& =& k-y^2\\ (x+c{)}^2+y^2& =& k\end{array}$

Is the required solution.

conclusion

The solution is

$(x+c{)}^2+y^2=k$